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Q.

Figure shows two cylinders of radii r1 and r2 having moments of inertia I1 and I2 about their respective axes. Initially, the cylinders rotate about their axes with angular speed ω1 and ω2, as shown in the figure. The cylinders are moved closer to touch each other keeping the axes parallel. The cylinders first slip over each other at the contact but the slipping finally ceases due to the friction between them. Find the angular speeds of the cylinders after the slipping ceases.

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a

ω1' =2 I1ω1r2 + I2ω2r1I2r12 + I1r22 r2, ω2' =2 I1ω1r2 + I2ω2r1I2r12 + I1r22 r1

b

ω1' = I1ω1r2 + I2ω2r1I2r12 + I1r22 r2,  ω2' = I1ω1r2 + I2ω2r1I2r12 + I1r22 r1

c

ω1' = I1ω1r2 + I2ω2r1I2r12 + I1r22 r1,   ω2' = I1ω1r2 + I2ω2r1I2r12 + I1r22 r2

d

ω1' = I2ω1r2 + I1ω2r1I2r12 + I1r22 r2, ω2' = I2ω1r2 + I1ω2r1I2r12 + I1r22 r1

answer is B.

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Detailed Solution

In the process, frictional force at contact point consitutes torque, which causes change in angular momentum of the cylinders. If ω1' and ω2' are the final angular velocities of the cylinder then for no
slipping

                         ω1' r1 =ω2' r2             ......(i)

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For first cylinder            f r1 t = I1 ω1'-ω1          .....(ii)

For second cylinder       f r2 t = I2 ω2'-ω2          .....(iii)

After solving above equations, we get 

             ω1' = I1ω1r2 + I2ω2r1I2r12 + I1r22 r2

and     ω2' = I1ω1r2 + I2ω2r1I2r12 + I1r22 r1

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